Consider the map from $\mathbb{C}$ to a unital Banach algebra $B$ given by $x \mapsto \exp(xb)$ for $b\in B$. I proved that this map is continuous by using the definition of $\exp(xb)$ as a contour integral around the spectrum of $b$.
However, I am wondering if there is an easier way to prove this without using the integral expression, for example by using the fact that the exponential map is continuous as a map $\mathbb{C}\rightarrow \mathbb{C}$, and/or by using some properties of the algebra of holomorphic functions on A? I have tried using the composition $x \mapsto xb \mapsto \exp(xb)$, but without succes...
Answered by Daniel Fischer: $$ \sum_{k=0}^\infty \frac{x^kb^k}{k!}$$ is a locally uniformly convergent series, hence the sum is continuous. Indeed, $$\lVert x^kb^k\rVert \leqslant \lvert x\rvert^k\lVert b\rVert^k$$ for every $k$. Since the exponential series converges everywhere in $\mathbb C$, the convergence is uniform in $|x|\le R$ for all $R<\infty$.